Question

Asteroid is moving directly towards the centre of the earth. When at a distance of $10R,$ ($R$ is the radius of the earth) from the earth's centre, it has a speed of $12\text{km/s}$. Neglecting the effect of earth atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is $11.2\text{km/s}$? Give your answer to the nearest integer in $\text{kmps}$

Answer 1

Using law of conservation of energy
Total energy at height $10R=$total energy at earth
$-\frac{G{M}_{E}m}{10R}+-\frac{1}{2}m{v}_0^2=-\frac{G{M}_{E}m}{R}+-\frac{1}{2}m{v}^2$
$[\because \text{Gravitational potential energy =}-\frac{GMm}{r}]$
$\Rightarrow \frac{G{M}_E}{R}(1-\frac{1}{10})+\frac{v_0^2}{2}=\frac{v^2}{2}$
$\Rightarrow v^2=v_0^2+\frac{9}{5}gR$
Given $[v_0=\text{12 km/s}]$
$\Rightarrow v=\sqrt{v_0^2+\frac{9}{5}gR}\approx 16\text{km/s}$